Question

The capacitors have values C1 = 2.0 μF and C2 = 4.0 μF, C3 = 5.0 μF, 

C4 = 7.0 μF and the potential difference across the battery is 9.0 V.

Assume that the capacitors are connected in series.

a) Find the equivalent capacitance of the circuit.

b) Solve for the potential difference across each capacitor.

Answer 1

Answer

a) equivalent capacitance is

$C=\frac{2*4*5*7}{4*5*7+2*5*7+2*4*7+2*4*5}\\=\frac{280}{306}\\=0.92\mu F$

b) potential difference is calculated by

Q=CV.